Given:

The points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)

**(i) **xy-plane

We know z = 0 in xy-plane.

So let P(x, y, 0) be any point in xy-plane

According to the question:

PA = PB = PC

PA^{2} = PB^{2} = PC^{2}

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

We know PA^{2} = PB^{2}

So, (x – 1)^{2}+ (y + 1)^{2} = (x – 2)^{2} + (y – 1)^{2} + 4

x^{2}+ 1 – 2x + y^{2} + 1 + 2y = x^{2}+ 4 – 4x + y^{2} + 1 – 2y + 4

– 2x + 2 + 2y = 9 – 4x – 2y

– 2x + 2 + 2y – 9 + 4x + 2y = 0

2x + 4y – 7 = 0

2x = – 4y + 7……………………(1)

Since, PA^{2} = PC^{2}

So, (x – 1)^{2}+ (y + 1)^{2} = (x – 3)^{2} + (y – 2)^{2} + 1

x^{2}+ 1 – 2x + y^{2} + 1 + 2y = x^{2}+ 9 – 6x + y^{2} + 4 – 4y + 1

– 2x + 2 + 2y = 14 – 6x – 4y

– 2x + 2 + 2y – 14 + 6x + 4y = 0

4x + 6y – 12 = 0

2(2x + 3y – 6) = 0

Now substitute the value of 2x (obtained in equation (1)), we get

7 – 4y + 3y – 6 = 0

– y + 1 = 0

y = 1

By substituting the value of y back in equation (1) we get,

2x = 7 – 4y

2x = 7 – 4(1)

2x = 3

x = \( \dfrac{3}{2}\)

The point P (\( \dfrac{3}{2}\), 1, 0) in xy-plane is equidistant from A, B and C.

**(ii) **yz-plane

We know x = 0 in yz-plane.

Let Q(0, y, z) any point in yz-plane

According to the question:

QA = QB = QC

QA^{2} = QB^{2} = QC^{2}

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

We know, QA^{2} = QB^{2}

So, 1 + z^{2}+ (y + 1)^{2} = (z – 2)^{2} + (y – 1)^{2} + 4

z^{2}+ 1 + y^{2} + 1 + 2y

= z^{2}+ 4 – 4z + y^{2} + 1 – 2y + 4

2 + 2y = 9 – 4z – 2y

2 + 2y – 9 + 4z + 2y = 0

4y + 4z – 7 = 0

4z = –4y + 7

z = \(\dfrac{ [–4y + 7]}{4}\) …. (1)

Since, QA^{2} = QC^{2}

So, 1 + z^{2}+ (y + 1)^{2} = (z + 1)^{2} + (y – 2)^{2} + 9

^{2}+ 1 + y^{2} + 1 + 2y = z^{2}+ 1 + 2z + y^{2} + 4 – 4y + 9

2 + 2y = 14 + 2z – 4y

2 + 2y – 14 – 2z + 4y = 0

–2z + 6y – 12 = 0

2(–z + 3y – 6) = 0

Now, substitute the value of z [obtained from (1)] we get

12y + 4y – 7 – 24 = 0

16y – 31 = 0

y = \( \dfrac{31}{16}\)

Substitute the value of y back in equation (1), we get

= \( \dfrac{-3}{16}\)

The point Q (0, \( \dfrac{31}{16}\), \( \dfrac{-3}{16}\)) in yz-plane is equidistant from A, B and C.

**(iii) **zx-plane

We know y = 0 in xz-plane.

Let R(x, 0, z) any point in xz-plane

According to the question:

RA = RB = RC

RA^{2} = RB^{2} = RC^{2}

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

We know, RA^{2} = RB^{2}

So, 1 + z^{2}+ (x – 1)^{2} = (z – 2)^{2} + (x – 2)^{2} + 1

z^{2}+ 1 + x^{2} + 1 – 2x = z^{2}+ 4 – 4z + x^{2} + 4 – 4x + 1

2 – 2x = 9 – 4z – 4x

2 + 4z – 9 + 4x – 2x = 0

2x + 4z – 7 = 0

2x = –4z + 7……………………………(1)

Since, RA^{2} = RC^{2}

So, 1 + z^{2}+ (x – 1)^{2} = (z + 1)^{2} + (x – 3)^{2} + 4

z^{2}+ 1 + x^{2} + 1 – 2x

= z^{2}+ 1 + 2z + x^{2} + 9 – 6x + 4

2 – 2x = 14 + 2z – 6x

2 – 2x – 14 – 2z + 6x = 0

–2z + 4x – 12 = 0

2(2x) = 12 + 2z

Substitute the value of 2x [obtained from equation (1)] we get,

2(–4z + 7) = 12 + 2z

–8z + 14 = 12 + 2z

14 – 12 = 8z + 2z

10z = 2

z = \( \dfrac{2}{10}\)

= \( \dfrac{1}{5}\)

Now, substitute the value of z back in equation (1), we get

2x = -4z + 7

The point R (\( \dfrac{31}{10}\), 0, \( \dfrac{1}{5}\)) in xz-plane is equidistant from A, B and C.

Answered by Aaryan | 1 year agoA(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z = 5.

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.